Match List-$I$ with List-$II$:
List-$I$ List-$II$
$A$. Gauss's Law in Electrostatics $I$. $\oint \vec{E} \cdot d \vec{l} = -\frac{d \phi_B}{d t}$
$B$. Faraday's Law $II$. $\oint \vec{B} \cdot d \vec{A} = 0$
$C$. Gauss's Law in Magnetism $III$. $\oint \vec{B} \cdot d \vec{l} = \mu_0 i_C + \mu_0 \epsilon_0 \frac{d \phi_E}{d t}$
$D$. Ampere-Maxwell Law $IV$. $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$

Choose the correct answer from the options given below:

  • A
    $A-IV, B-I, C-II, D-III$
  • B
    $A-I, B-II, C-III, D-IV$
  • C
    $A-III, B-IV, C-I, D-II$
  • D
    $A-II, B-III, C-IV, D-I$

Explore More

Similar Questions

According to Maxwell,which is the missing term in Maxwell-Ampere circuital law?

If the rate of change in electric flux between the plates of a capacitor is $9 \pi \times 10^3 \text{ Vm s}^{-1}$,then the displacement current inside the capacitor is (in $\mu \text{A}$)

Show that the magnetic field $B$ at a point in between the plates of a parallel-plate capacitor during charging is $B = \frac{{\mu _0 \epsilon _0 r}}{2} \cdot \frac{{dE}}{{dt}}$ (symbols have their usual meanings).

"Changing electric field produces magnetic field". Explain the importance of the given statement.

$A$ parallel plate capacitor made of circular plates is being charged such that the surface charge density on its plates is increasing at a constant rate with time. The magnetic field arising due to displacement current is

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo